Metamath Proof Explorer


Theorem rngqiprngfulem3

Description: Lemma 3 for rngqiprngfu (and lemma for rngqiprngu ). (Contributed by AV, 16-Mar-2025)

Ref Expression
Hypotheses rngqiprngfu.r φ R Rng
rngqiprngfu.i φ I 2Ideal R
rngqiprngfu.j J = R 𝑠 I
rngqiprngfu.u φ J Ring
rngqiprngfu.b B = Base R
rngqiprngfu.t · ˙ = R
rngqiprngfu.1 1 ˙ = 1 J
rngqiprngfu.g ˙ = R ~ QG I
rngqiprngfu.q Q = R / 𝑠 ˙
rngqiprngfu.v φ Q Ring
rngqiprngfu.e φ E 1 Q
rngqiprngfu.m - ˙ = - R
rngqiprngfu.a + ˙ = + R
rngqiprngfu.n U = E - ˙ 1 ˙ · ˙ E + ˙ 1 ˙
Assertion rngqiprngfulem3 φ U B

Proof

Step Hyp Ref Expression
1 rngqiprngfu.r φ R Rng
2 rngqiprngfu.i φ I 2Ideal R
3 rngqiprngfu.j J = R 𝑠 I
4 rngqiprngfu.u φ J Ring
5 rngqiprngfu.b B = Base R
6 rngqiprngfu.t · ˙ = R
7 rngqiprngfu.1 1 ˙ = 1 J
8 rngqiprngfu.g ˙ = R ~ QG I
9 rngqiprngfu.q Q = R / 𝑠 ˙
10 rngqiprngfu.v φ Q Ring
11 rngqiprngfu.e φ E 1 Q
12 rngqiprngfu.m - ˙ = - R
13 rngqiprngfu.a + ˙ = + R
14 rngqiprngfu.n U = E - ˙ 1 ˙ · ˙ E + ˙ 1 ˙
15 rnggrp R Rng R Grp
16 1 15 syl φ R Grp
17 1 2 3 4 5 6 7 8 9 10 11 rngqiprngfulem2 φ E B
18 1 2 3 4 5 6 7 rngqiprng1elbas φ 1 ˙ B
19 5 6 rngcl R Rng 1 ˙ B E B 1 ˙ · ˙ E B
20 1 18 17 19 syl3anc φ 1 ˙ · ˙ E B
21 5 12 grpsubcl R Grp E B 1 ˙ · ˙ E B E - ˙ 1 ˙ · ˙ E B
22 16 17 20 21 syl3anc φ E - ˙ 1 ˙ · ˙ E B
23 5 13 16 22 18 grpcld φ E - ˙ 1 ˙ · ˙ E + ˙ 1 ˙ B
24 14 23 eqeltrid φ U B